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Engineering

Goodman, Soderberg and Gerber Proportional-Loading Factors

Compare three mean-stress fatigue factors, calculate the Gerber root stably and check the peak-stress yield factor separately.

State what the factor scales

For a uniaxial cycle, amplitude sigma_a=(sigma_max-sigma_min)/2 is nonnegative and mean sigma_m=(sigma_max+sigma_min)/2 can have either sign. Enter stress and strengths in the same unit, here MPa. Se must be a suitable corrected endurance limit, or an explicitly chosen fully reversed fatigue strength for the target life. Raw polished-specimen data is not automatically a component strength.

These factors scale both the alternating and mean stress by n along a constant-ratio loading line. They do not hold bolt preload or another fixed mean load constant while increasing only the amplitude. A fixed-mean loading path needs a different calculation.

Compare the straight-line criteria

E4 computes modified Goodman: n=1/(sigma_a/Se+mean_used/Sut). E5 computes Soderberg by replacing Sut with Sy in the mean term. The initial data gives Goodman1.25 and Soderberg1.111111…. These are model factors, not a certification of safe operation.

E2 requires numeric inputs, nonnegative amplitude, positive strengths and Sy<=Sut. Soderberg is more restrictive than Goodman for tensile mean stress under that strength ordering. Do not assume either model covers every fatigue mechanism.

=IF(E2<>"OK",E2,IF(B2+E3=0,"No finite fatigue factor",1/(B2/B4+E3/B5)))

Calculate the missing Gerber factor

Let a=sigma_a/Se and b=mean_used/Sut. Proportional loading reaches the Gerber curve when n*a+(n*b)^2=1. The positive root can be written n=2/(a+SQRT(a^2+4*b^2)). This avoids subtracting nearly equal terms and also works for b=0 when a>0.

E6 evaluates that expression directly. The initial result is about1.561250. It differs from 1/(a+b^2), which does not scale both stress components according to the quadratic equation. Keep raw outputs and round only for display.

=IF(E2<>"OK",E2,IF(B2+E3=0,"No finite fatigue factor",2/(B2/B4+SQRT((B2/B4)^2+4*(E3/B5)^2))))

Make the compression assumption explicit

E3 uses mean_used=MAX(0,sigma_m) for the three fatigue calculations. This is a deliberate no-benefit treatment of compressive mean stress, not a universal rule that compression cannot cause failure. With sigma_m=-50 and sigma_a=140, each fatigue model returns2.

For zero amplitude and a nonpositive mean, the fatigue denominator is zero: the sheet reports No finite fatigue factor. It does not declare the component safe. In particular, a large static compressive load still needs strength and stability checks.

Check peak stress and yielding separately

For this scalar uniaxial cycle, E7=ABS(sigma_m)+sigma_a is the greatest absolute stress. E8=Sy/E7 gives a separate yield factor, assuming the same yield strength in tension and compression. The initial peak is320 MPa and the yield factor is1.40625, which is below the Gerber fatigue factor.

Zero total stress returns No applied stress. A cycle with zero amplitude is static: the fatigue expressions merely reach their mathematical intercepts. Assess static failure separately. Soderberg only implies the simple tensile-side yield check under additional assumptions such as Se<=Sy; the sheet does not rely on that implication.

=IF(E2<>"OK",E2,IF(E7=0,"No applied stress",B6/E7))

Try cases and interpret the limits

Set mean stress to0: all three fatigue factors become Se/sigma_a=2. Set mean stress to-500: the no-benefit fatigue factor stays2, but the yield factor falls to450/640=0.703125. Restore the inputs, then double amplitude and mean together: every factor halves.

A factor below1 indicates crossing the corresponding model boundary; a factor above1 does not prove adequate design margin. Choose the required factor and material model separately. Multiaxial stresses, notches, local plasticity, variable-amplitude damage, surface finish, environment and component-specific failure modes need additional assessment. Materials without a defensible endurance limit should not be labeled infinite-life solely from these formulas.

Try the example

Use the original values or change them to test your understanding.

Further reading